Question 105 - In a rigid rotor model in quantum chemistry, the moment of inertia I is given by an Equation E as I = Ma x La x La + Mc x Lc x Lc = m x L x L, where m = (Ma x Mc) / (Ma + Mc) and L = La + Lc, m is the reduced mass, Ma is the mass of a, Mc is the mass of c, La is the radius of a from point O, Lc is the radius of c from point O. Prove by simplest method that Equation E is wrong.
Answer 105 - Let Ma = 0 and Mc = 1 as assumption. Substitute them into Equation E as I = Ma x La x La + Mc x Lc x Lc = 0 + Lc x Lc = Lc x Lc. However, m = (0 x 1) / (0 + 1) = 0, then I = Lc x Lc, which is not equal to 0 but I = m x L x L = 0 x L x L = 0. Equation E is proven wrong. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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Question 33 – By using Excel program either on laptop or desktop PC, solve the differential equation dy / dx = -2y + x + 4 with h = 0.005, initial values : x = 0, y = 1. The 4th order Runge-Kutta method provides : y(N + 1) = y(N) + (1/6) (k1 + 2k2 +2k3 + k4), k1 = h [ -2y(N) + x(N) + 4 ], k2 = h { -2 [ y(N) + k1 / 2 ] + x(N) + h / 2 + 4 }, k3 = h { -2 [ y(N) + k2 / 2 ] + x(N) + h / 2 + 4 }, k4 = h { -2 [ y(N) + k3 ] + x(N) + h + 4 }. What is the value of y at x = 0.5?
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DIFFERENTIAL EQUATIONS - EXAMPLE 20.1 : By using Excel program either on laptop or desktop PC, solve the differential equation dy / dx = -2y + x + 4 with h = 0.005, initial values : x = 0, y = 1. The 4th order Runge-Kutta method provides : y(N + 1) = y(N) + (1/6) (k1 + 2k2 +2k3 + k4), k1 = h [ -2y(N) + x(N) + 4 ], k2 = h { -2 [ y(N) + k1 / 2 ] + x(N) + h / 2 + 4 }, k3 = h { -2 [ y(N) + k2 / 2 ] + x(N) + h / 2 + 4 }, k4 = h { -2 [ y(N) + k3 ] + x(N) + h + 4 }. What is the value of y at x = 0.5?
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